What is the volume when the region is revolved about the x-axis

the region is bounded by y = 4 - x^2, the x-axis, and the y-axis.

i tried 512pi/15 and that was wrong. would it be 256pi/15? or something else?

The axis of rotation is horizontal, so the washer method will require us to slice vertically, implying dx. Note that the limits will extend from 0 to 2:

V = ∫ A(x) dx from 0 to 2

A(x) = πR^2 - πr^2 = π(R^2 - r^2)

So first determine the outer radius, R. The radius is defined by the parabola, so R = 4 - x^2. Next determine the inner radius, r. The inner radius is defined by the line y = 0, so r = 0. Sub those in:

A(x) = π([4 - x^2]^2 - 0^2)
A(x) = π(x^4 - 8x^2 + 16)

V = π ∫ [x^4 - 8x^2 + 16] dx from 0 to 2
V = π[(1/5)x^5 - (8/3)x^3 + 16x] from 0 to 2

Now evaluate and note that the lower limit of 0 will be 0:

V = π[(1/5)(2)^5 - (8/3)(2)^3 + 16(2)]
V = (256/15)π

Done!